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          <h1 class="post-title" itemprop="name headline">Python Pandas 构建共现矩阵</h1>
        

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        <blockquote>
<p>作者共现、词共现等各种共现情况的研究需求在我们这个学科还是挺多的，来写一下如何使用Python构建共现矩阵。</p>
</blockquote>
<a id="more"></a>
<h3 id="基本思路"><a href="#基本思路" class="headerlink" title="基本思路"></a>基本思路</h3><blockquote>
<p>构建共现矩阵，我一个基本思路是先根据所有的目标对象构建一个单位矩阵，然后对目标对象的共现情况进行统计、计算，将相应的结果覆盖到对应的矩阵元素即可。对于这种矩阵来说，矩阵的索引、列名使用对应的目标对象的名称会比较方便，因而想到了使用Python中的Pandas来构建矩阵。</p>
</blockquote>
<h3 id="目标对象统计"><a href="#目标对象统计" class="headerlink" title="目标对象统计"></a>目标对象统计</h3><blockquote>
<p>以作者共现为例。首先对所有合作者的基本情况进行统计，分别构建一个字典统计所有作者单独出现的次数和一个字典统计所有作者两两出现的次数。</p>
</blockquote>
<ul>
<li>将所有的合作者名字放入一个列表，以逗号隔开。</li>
<li>单个作者出现的频次，按照一般频次统计的方法，判断该对象是否在字典中，不在则对应的value初始化为1，在则value加1。</li>
<li>确定当前作者的合作者，目前的一个基本想法是遍历合作者样本，对于每一个作者来说，其合作者是排在他后面的所有作者。于是复制一个样本，每次遍历时去掉第一个也就是当前的作者，那么这个复制样本中所有的作者即为当前作者的合作者。（这个思路其实也是一个组合算法，不知道还有没有优化的方法😔，，还请指点）</li>
<li>将这两个合作者以逗号连接，用同样的方法统计频次。但是要注意一个问题，每个样本中两个作者的顺序可能不一致，因此要进行判断，统一顺序。</li>
<li>（2018.06.30修改）之前的代码有一个错误，上面的统一顺序要增加两个变量来替换au和au_c的值，因为au位于循环中，不能改变其本身的值。</li>
</ul>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">def</span> <span class="title">authors_stat</span><span class="params">(co_authors_list)</span>:</span></span><br><span class="line">    au_dict = &#123;&#125;  <span class="comment"># 单个作者频次统计</span></span><br><span class="line">    au_group = &#123;&#125;  <span class="comment"># 两两作者合作</span></span><br><span class="line">    <span class="keyword">for</span> authors <span class="keyword">in</span> co_authors_list:</span><br><span class="line">        authors = authors.split(<span class="string">','</span>)  <span class="comment"># 按照逗号分开每个作者</span></span><br><span class="line">        authors_co = authors  <span class="comment"># 合作者同样构建一个样本</span></span><br><span class="line">        <span class="keyword">for</span> au <span class="keyword">in</span> authors:</span><br><span class="line">            <span class="comment"># 统计单个作者出现的频次</span></span><br><span class="line">            <span class="keyword">if</span> au <span class="keyword">not</span> <span class="keyword">in</span> au_dict:</span><br><span class="line">                au_dict[au] = <span class="number">1</span></span><br><span class="line">            <span class="keyword">else</span>:</span><br><span class="line">                au_dict[au] += <span class="number">1</span></span><br><span class="line">            <span class="comment"># 统计合作的频次</span></span><br><span class="line">            authors_co = authors_co[<span class="number">1</span>:]  <span class="comment"># 去掉当前作者</span></span><br><span class="line">            <span class="keyword">for</span> au_c <span class="keyword">in</span> authors_co:</span><br><span class="line">                A, B = au, au_c  <span class="comment"># 不能用本来的名字，否则会改变au自身</span></span><br><span class="line">                <span class="keyword">if</span> A &gt; B:</span><br><span class="line">                    A, B = B, A  <span class="comment"># 保持两个作者名字顺序一致</span></span><br><span class="line">                co_au = A + <span class="string">','</span> + B  <span class="comment"># 将两个作者合并起来，依然以逗号隔开</span></span><br><span class="line">                <span class="keyword">if</span> co_au <span class="keyword">not</span> <span class="keyword">in</span> au_group:</span><br><span class="line">                    au_group[co_au] = <span class="number">1</span></span><br><span class="line">                <span class="keyword">else</span>:</span><br><span class="line">                    au_group[co_au] += <span class="number">1</span></span><br><span class="line">    <span class="keyword">return</span> au_group, au_dict</span><br></pre></td></tr></table></figure>
<h3 id="构建单位矩阵"><a href="#构建单位矩阵" class="headerlink" title="构建单位矩阵"></a>构建单位矩阵</h3><p>利用Pandas，以作者名为索引和列表，构建一个单位矩阵：</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">import</span> pandas <span class="keyword">as</span> pd</span><br><span class="line"><span class="keyword">import</span> numpy <span class="keyword">as</span> np</span><br><span class="line">au_list = list(au_dict.keys())  <span class="comment"># 取出所有单个作者</span></span><br><span class="line">matrix = pd.DataFrame(np.identity(len(au_list)), columns=au_list, index=au_list)</span><br></pre></td></tr></table></figure>
<h3 id="共现系数计算"><a href="#共现系数计算" class="headerlink" title="共现系数计算"></a>共现系数计算</h3><p>共现矩阵中的元素采用Equivalence系数进行归一化，公式如下：<br>$$<br>E_{ij}={F_{ij}^2 \over F_i*F_j}<br>$$<br>其中，Eij为共现矩阵元素的值，Fij为两个目标对象i和j共现的字数，Fi为目标对象i出现的总频次，Fj为目标对象j出现的总频次。</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">def</span> <span class="title">generate_matrix</span><span class="params">(au_group, matrix)</span>:</span></span><br><span class="line">    <span class="keyword">for</span> key, value <span class="keyword">in</span> au_group.items():</span><br><span class="line">        A = key.split(<span class="string">','</span>)[<span class="number">0</span>]</span><br><span class="line">        B = key.split(<span class="string">','</span>)[<span class="number">1</span>]</span><br><span class="line">        Fi = au_dict[A]</span><br><span class="line">        Fj = au_dict[B]</span><br><span class="line">        Eij = value*value/(Fi*Fj)</span><br><span class="line">        <span class="comment">#按照作者进行索引，更新矩阵</span></span><br><span class="line">        matrix.ix[A, B] = Eij</span><br><span class="line">        matrix.ix[B, A] = Eij</span><br><span class="line">    <span class="keyword">return</span> matrix</span><br></pre></td></tr></table></figure>
<p>以上。</p>
<p>完整源代码==&gt;<a href="https://github.com/LeoWood/CoMatrix/blob/master/CoMatrix.py" target="_blank" rel="noopener">CoMatrix.py</a></p>

      
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